1. The density at $20^\circ\mathrm{C}$ of a $\pu{0.500 M}$ solution of acetic acid in water is $\pu{1.0042 g/mL}$. What is the molality of the solution? The molar mass of acetic acid is $M(\ce{CH3CO2H})=\pu{60.05 g/mol}$.
  2. The density at $20^\circ\mathrm{C}$ of a $\pu{0.258 M}$ solution of glucose in water is $\pu{1.0173 g/mL}$, and the molar mass of glucose is $M(\ce{Glc}) = \pu{180.2 g/mol}$. What is the molarity of the solution?

In question one we can assume a 1 liter solution and in question two we can assume a 100 gram sample. I understand these assumptions are needed to find other information, but when are we justified in making these assumptions?


1 Answer 1


You can make certain assumptions in these and similar problems because concentration of uniform substances (whether it's expressed as molarity, molality, mass density, or something else entirely) is an intensive property, i.e., it has no dependence on the total amount of material in the object or sample being measured. For example, if you have a homogeneous solution, the solute particles are distributed uniformly among the particles of solvent, and the two form a constant ratio. If I split the sample in half, the solute particles are halved, but so are the solvent particles, so the ratio remains the same.

As it happens, there's no need to assume any particular quantities to solve this problem; instead, you can use algebraic substitutions using the known values given and common definitions to solve for either molarity or molality directly. Let $M = \frac{m_{solute}}{n_{solute}}$ be the molar mass of the solute, $\rho = \frac{m_{solution}}{V_{solution}}$ be the mass density of the solution, $c = \frac{n_{solute}}{V_{solution}}$ be the molarity, and $b = \frac{n_{solute}}{m_{solvent}}$ be the molality, and note that $m_{solvent} = m_{solution} - m_{solute}$. Then, we have:

$$ \begin{align} b\ & = \frac{n_{solute}}{m_{solution} - m_{solute}}\\ & = \frac{n_{solute}}{\rho \cdot V_{solution} - M \cdot n_{solute}}\\ & = \frac{n_{solute}}{\rho \cdot \frac{n_{solute}}{c} - M \cdot n_{solute}}\\ & = \frac{c}{\rho - M \cdot c}\\ \end{align} $$

You can rearrange this equation to solve for $c$ (molarity). Note, of course, that units need to match, but it's trivial to convert between grams and kilograms, mililiters and liters, etc.


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